February 21, 2022
… why isn’t it perfect?
What are some potential sources of “randomness” in the sea otter population process?
Easiest problem in boring arithmetic: \[\huge 1 + 1 = 2\]
\(\huge +\) \(\huge = ?\)
Random variables have possible events associated with probabilities.
| event | numeric | probability |
|---|---|---|
 |
\(\Large 0\) | \(\Large 1/2\) |
 |
\(\Large 1\) | \(\Large 1/2\) |
Random variables have possible events associated with probabilities.
| event | numeric | probability |
|---|---|---|
|
\(\Large 0\) | \(\Large 1/4\) |
|
\(\Large 1\) | \(\Large 1/2\) |
|
\(\Large 2\) | \(\Large 1/4\) |
… are described by Probability Distributions. Probability Distributions have names and parameters that describe the distribution.
\(X \sim Binomial(n, p)\) describes the sum of \(n\) random events, each of which has probability \(p\).
If you have 20 sea otters (\(X_t = 20\)), each with a 90% probability of survival / year, how many next year?
Answer: A random variable with distribution: \(X_{t+1} \sim Binomial(n = 20, p = 0.9)\)
\[X_{t+1} \sim Binomial(n = X_t,\,\, p = p_{surv})\]
\[E(X) = \mu_X = np = 0.9 \times 20 = 18\]
This value is the Expectation of the distribution: \(E(X)\) or \(\mu(X)\). It would be the mean value if you could repeat the experiment an infinite amount of times.
\[SD(X) = \sigma_x = \sqrt{n\,p(1-p)} = 1.34\] The standard deviation of a random variable: \(SD(X)\) or \(\sigma_x\) quantifies how concentrated the distribution is around the mean. Approximately 95% of the probability is within 2 standard deviations.
There are a bunch. The most famous is the Normal or Gaussian distribution:
\[X \sim {\cal N}(\mu, \sigma)\]
It has two parameters: \(\mu\) and \(\sigma\) with are - unsurprisingly - the mean (expectation) and standard deviation (spread) of the variable.
We use these to model random processes:
| name | notation | possible values | models | mean | standard deviation |
|---|---|---|---|---|---|
| Normal | \({\cal N}(\mu, \sigma)\) | \((-\infty, \infty)\) | bell-shaped | \(\mu\) | \(\sigma\) |
| Exponential | \(Exp( \lambda)\) | \([0,\infty)\) | random events | \({1\over\lambda}\) | \({1 \over \lambda}\) |
| Poisson | \(Poisson(\lambda)\) | \([0,\infty)\) | positive count data (e.g. births) | \(\lambda\) | \(\sqrt{\lambda}\) |
| Bernoulli | \(Bernoulli(p)\) | \([0,1)\) | binary outcomes (e.g. deaths) | \(p\) | \(\sqrt{p(1-p)}\) |
| Binomial | \(Binomial(n, p)\) | \([0,n)\) | many binary outcomes | \(np\) | \(\sqrt{n p(1-p)}\) |
Stochasticity means: Randomness in time.
Demography is the Science of Population Dynamics. Often it refers specifically to births and deaths (and movements … but we’re still looking at closed population).
Individually, all demographic processes are stochastic. An individual has some probability of dying at any moment. An individual has some probability of reproducting at a given time.
Q: How important is individual randomness for a population process?
More specific Q: What is the probability of extinction?
Starting with \(N_t\): expected number of survivors (S): \[E(S) = p_s N_t\] Expected number of new individuals (babies - B): \[E(B) = p_b E(S)\]
New population equals survivals + new babies:
\[E(N_{t+1}) = E(S) + E(B) = p_s N_t + p_b\,p_s N_t = p_s(1 + p_b) N_t\] So (in our coin flip example) \[\widehat{\lambda} = p_s (1+p_b) = 0.75\].
https://egurarie.shinyapps.io/StochasticGrowth/
(… for a similar continuous model).
Assume birth rate \(b\) and death rate \(d\), and growth rate \(r = b-d\).
The mean of the process is also exponential growth
\[E(N_t) = N_0 e^{r t}\] If birth rate = death rate:
\[SD(N_t) \approx \sqrt{2 N_0 b t}\]
Note, increases as square root of time. There’s a somewhat more complex formula for births \(\neq\) deaths.
More importantly:
\[ P(extinction) = \begin{cases} \text{if}\,\,b > d;& (d/b)^{N_0}\\ \text{if} \,\, d > b;& 1 \end{cases}\] Note that even when birth rate = death rate, \(P(extinction) = 1\), i.e. eventual extinction is certain. This is very similar to the eventual probability of fixation for genetic drift.
Also, probability of extinction is lower for smaller \(N_0\).
… refers to some random aspect of the environment affecting \(r\) (whether via births or deaths or both) for the entire population. :
\[R \sim Dist(\mu_r, \sigma_r)\]
\(N_0 = 16\)
If I (the environment) flip Heads, the population doubles.
If I flip Tails, the population halves.
This model can be written:
\[N_{t+1} = N_t \, 2^{R}\]
Where \(R\) is a random variable:
| value | prob. |
|---|---|
| -1 | 1/2 |
| +1 | 1/2 |
\[E(R) = \mu_r = 0\]
So population should not grow, on average.
\[E(N(r)) = N_0 e^{\mu_r t} = N_0\]
And:
\[SD(R) = \sigma_R = 1\]
Are you doomed to extinction? Does this remind you of another process from earlier in class?
Experiment here: https://egurarie.shinyapps.io/StochasticGrowth
\[N(t) \sim Dist(\mu_N(t), \sigma_N(t))\]
The mean / expectation is the same as for the exponential growth function, but with the mean growth rate substituted:
\[\mu_N(t)= N_0 e^{\mu_r t}\] But the standard deviation of the function increases with time!
\[\sigma^2_{N}(t) \approx N_0\,e^{\mu_r t} \sqrt{e^{\sigma_r^2t} - 1}\]
Sea otter example, with \(\mu_r = 0.07\) and \(\sigma_r = 0.07\):
Grey lines are simulations. Red lines are \(\mu_N(t) \pm \sigma_N(t)\)
that spells trouble for a population!
According to theory, if \(\sigma_r > 2\mu_r\), extinction is nearly certain - even if \(\mu_r\) is positive and on average there is growth.
Opportunities for experimentation: